B Why is "time orthogonal to space" in inertial reference frames?

[pervect]

Apologies in advance if I'm addressing points already raised, I've only skimmed the thread.

Experimentally, time being orthogonal to space is the same as saying that we use Einstein clock synchronization convention, as a way to get "fair" or "isotropic" clock synchronization.

When we make time orthogonal to space, physics work the way we usually expect. Light moves the same in both directions, and so do electron beams with the same energy. It doesn't matter to the physics which way the light or the electron beam is going, it's speed is the same.

If we don't make time orthogonal to space, both the speed of light and the speed of electron beams of a specified energy depends on the direction. This is most obvious for high energy electron beams, as they approach the speed of light as one increases their energy per electron. Thus if the behavior of light is not isotropic, neither is the behavior of other things. See for instance and/or the peer reviewed paper associated with the video. The general topic is known as "isotropy". Sometimes people appear to think isotropy applies only to light, but it applies to other physical phenomenon as well.

If the idea that the speed of identical objects (identical because they have a known energy, or momentum) moving in different directitons is uncomfortable, then one should stick with isotropic coordinates and/or inertial frames, which implies that time must be orthogonal to space.

 

[PeterDonis]

Linearity of the transformation follows from the fact that in IRFs free particles have constant speed.

Yes.

Assuming that the new chart $(t',x')$ representing another IRF is also rectilinear

I don't think this assumption for the new chart is actually required. If you look at coordinate speeds in non-rectilinear charts, you will see that they can't possibly be constant in all directions. For example, consider $d\theta / dt$ for a free particle in spherical coordinates (and ignore the edge case where the particle happens to be moving along an axis that makes $\theta$ constant). So constant coordinate speed in all directions already implies a rectilinear chart.

hopefully this is an acceptable physical justification of the homogeneity and isotropy assumptions.

It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.

I should've said that spacelike (timelike) basis vectors are mapped to spacelike (timelike) basis vectors in the new chart.

Meaning, mapped by the coordinate transformation? Yes, you can think of the transformation that way, but you can also think of it as simply changing the components of vectors, but not the vectors themselves. In other words, the old basis vectors will have more than one nonzero component in the new chart, and the new basis vectors will have more than one nonzero component in the old chart. The question then is how the coordinate transformation can change those components while still preserving the required properties for an IRF.

now I can only say that the images of the old chart's basis vectors are orthogonal.

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

This doesn't enable me to say anything about the new chart's basis vectors - whether both have the same or different signatures, or the inner product of those.

You can't assume the new basis vectors are orthogonal, or even that one is timelike and three are spacelike, that's true. You have to prove that, if it is provable.

You do have information about the new basis vectors, though: you know that free particles still have to have straight worldlines, and that light still has to have speed $c$. And you know that the coordinate transformation has to be linear. So you do have information to work from.

 

[Shirish]

...you know that free particles still have to have straight worldlines, and that light still has to have speed $c$. And you know that the coordinate transformation has to be linear. So you do have information to work from.

This info is used to derive the Lorentz transformation - in the $(t,x)\to(t',x')$ case, it's $\mathcal{L}=\begin{bmatrix}\gamma & \beta\gamma\\\beta\gamma & \gamma\end{bmatrix}$ (sign of off-diagonal terms depends on relative velocity so let's consider them positive in this case).

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

Now this is important info and something I'm embarrassed to admit that I didn't know. I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is (googling confirmed this). So if we have a couple of vectors $\mathbf{a},\mathbf{b}$ with representations $[ a ]_{\mathcal{A}}, [ b ]_{\mathcal{A}}$ in the initial coordinate chart $\mathcal{A}$ (and since we know that $\eta=\text{diag}(-1,1)$ is the metric representation in $\mathcal{A}$ by construction), their inner product is $[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}$.

Let $\eta'$ be the metric representation in the new coordinate chart $\mathcal{B}$ (the Minkowski metric doesn't change - its representation changes - something I realized after reading your post and googling related stuff). The representation of $\mathbf{a}$ in $\mathcal{B}$ will be $[ a ]_{\mathcal{B}}=\mathcal{L}[ a ]_{\mathcal{A}}$ (similarly for $\mathbf{b}$). Then
$$[ a ]^T_{\mathcal{B}}\eta'[ b ]_{\mathcal{B}}=[ a ]^T_{\mathcal{A}}\eta[ b ]_{\mathcal{A}}=[ a ]^T_{\mathcal{A}}\mathcal{L}^T(\mathcal{L}^T)^{-1}\eta\mathcal{L}^{-1}\mathcal{L}[ b ]_{\mathcal{A}}
=[ a ]^T_{\mathcal{B}}(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}[ b ]_{\mathcal{B}}
\\\implies \eta'=(\mathcal{L}^{-1})^T\eta\mathcal{L}^{-1}=\eta$$
So this was the missing link. And now since the metric representation is the same even in the new coordinate chart, and since the representation of the new basis vectors in the new chart $\mathcal{B}$ is $[1,0]$ and $[0,1]$, their inner product is zero.

It's a good start, but you spoke in general terms. There is a specific requirement for an IRF that requires isotropy.

I confess I don't know about that requirement. As far as I've read, isotropy is justified loosely by the fact that the orientation of the relative velocity vector between IRFs can be arbitrary. If there's an even stronger/more specific justification, I'm lost.

 

[PeroK]

Orthogonality is a coordinate-independent property, so a pair of vectors that are orthogonal in one chart will be orthogonal in any chart.

Although that itself does need to be proved at some stage.

 

[PeterDonis]

This info is used to derive the Lorentz transformation

How?

I thought if orthogonality is coordinate-independent as you said, then maybe even the inner product is

It is.

the metric representation is the same even in the new coordinate chart

It is if you use the Lorentz transformation, yes. But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).

I confess I don't know about that requirement.

Sure you do; we've mentioned it repeatedly in this discussion. It's even mentioned in the first quote from my previous post that you give in this post: look at the clause after the comma in the first sentence of that quote.

 

[PeterDonis]

that itself does need to be proved at some stage

It's obvious from the definition of orthogonality and the fact that inner products are coordinate-independent. And before you say that inner products being coordinate-independent has to be proved, that's part of the definition of a metric.

 

[Shirish]

But you still have to show how you derive the Lorentz transformation given the info you quoted at the top of your post (see my first question above).

One thing - assuming that the transformation is Lorentz, does the argument in my post (which shows $\eta=\eta'$ and hence orthogonality of basis in the new chart) seem okay to you?

As for Lorentz, there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz: https://www.physicsforums.com/threa...locks-are-synchronized-at-each-origin.989150/

The proof in the top post in that thread is incomplete, but folks at PF helped me complete it using homogeneity and isotropy assumptions (homogeneity to show that without loss of generality, we can assume a linear transformation instead of affine, and isotropy to show that $\gamma(v)$ is an even function of $v$).

My bad, I probably should've linked that earlier for some context.

 

[PeterDonis]

assuming that the transformation is Lorentz, does the argument in my post (which shows $\eta=\eta'$ and hence orthogonality of basis in the new chart) seem okay to you?

If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form $\eta$ of the metric.

there's a related thread here that I'd created previously in which I showed steps for showing that the transformation is Lorentz

Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being $c$ in the new coordinates, which is one of the requirements for an IRF.

 

[Shirish]

If the transformation is Lorentz, then yes, you're just expressing the well-known fact that the Lorentz transformation preserves the Minkowski form $\eta$ of the metric.
Ah, ok. Yes, if you use the derivation given in that thread, you're basically showing that the transformation has to be Lorentz in order to satisfy the speed of light being $c$ in the new coordinates, which is one of the requirements for an IRF.

Yay! So finally my mind is at peace. Now I can happily continue with the rest of the studies (until next time a doubt derails me again )

Thank you so much

 

[PeterDonis]

Thank you so much

You're welcome!